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soldi70 [24.7K]
3 years ago
9

The bottom of a box has a surface area of 25.0 cm2. The mass of the box is 34.0 kilograms. Acceleration due to gravity at sea le

vel is 9.80 m/s2. What pressure is exerted by the box where it rests? 1.36 N/cm2 13.3 N/cm2 0.139 N/cm2 7.21 N/cm2
Chemistry
2 answers:
oksano4ka [1.4K]3 years ago
6 0
Pressure is the force exerted per area in a system.  Force is equal to the mass times the acceleration. In this case, force is equal to the weight. We calculate pressure as follows:

P = F/A
P = mg/A
P = 34.0 (9.8) / 25.0
P = 13.3 N/cm^2

Therefore, the second option is the answer.
lyudmila [28]3 years ago
5 0
Pressure is given by:
Pressure=(Force)/(Area)
Force=mass*gravitational pull
mass=34 Kg
gravity=9.80
thus;
Force=34*9.8=333.2
thus;
pressure=333.2/25=13.328=13.3 N/cm^2
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3 years ago
How many grams of silver chloride can be produced by reacting excess silver nitrate with 2.4 moles of zinc chloride? _____AgNO3
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<h3>Answer:</h3>

690 g AgCl

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u>   </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Writing Compounds

<u>Stoichiometry</u>

  • Using Dimensional Analysis
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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Unbalanced] AgNO₃ + ZnCl₂ → AgCl + Zn(NO₃)₂

↓

[RxN - Balanced] 2AgNO₃ + ZnCl₂ → 2AgCl + Zn(NO₃)₂

[Given] 2.4 mol ZnCl₂

[Solve] <em>x</em> g AgCl

<u>Step 2: Identify Conversions</u>

[RxN] 1 mol ZnCl₂ → 2 mol AgCl

[PT] Molar Mass of Ag - 107.87 g/mol

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of AgCl - 107.87 + 35.45 = 143.32 g/mol

<u>Step 3: Stoich</u>

  1. [DA] Set up:                                                                                                      \displaystyle 2.4 \ mol \ ZnCl_2(\frac{2 \ mol \ AgCl}{1 \ mol \ ZnCl_2})(\frac{143.32 \ g \ AgCl}{1 \ mol \ AgCl})
  2. [DA] Multiply/Divide [Cancel out units]:                                                          \displaystyle 687.936 \ g \ AgCl

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

687.936 g AgCl ≈ 690 g AgCl

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Answer:

a) Representation - (in attachment)

b) Tetrahedral geometry and

c) sp^{3} hybrid orbitals invovle sigma bonding.

\pi orbitals are formed by the overlapping of d-orbitals of sulfur with p-orbitals of oxygen.

Explanation:

a)

Representation in attachment.

The overall charge in the both structures are -2. The structure (b) is favored by the resonance structures since the formal charge with in the species are remained.

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b)

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There is four sigma bonds and zero lone pairs present on the central metal atom.

Hence, the hybridization of the sulfur atom is sp^{3}

c)

The s and p orbitals of sulfur invovles hybridization and form sigma bonds.The orbitals of sulfur involves \pi bonding.

Therefore, \pi bonds are formed by the overlapping of d-orbitals of sulfur with  p-orbitals of oxygen.

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